Exercise 9, Chapter 2 of Stein's Fourier Analysis. Showing that a fourier series does not converge absolutely but converges conditionally.

Question

Let $f(x)=\chi_{[a,b]}(x)$ be the characteristic function of the interval $[a,b]\subset [-\pi,\pi]$.

Show that if $a\neq -\pi$, or $b\neq \pi$ and $a\neq b$, then the Fourier series does not converge absolutely for any $x$. [Hint: It suffices to prove that for many values of $n$ one has $|\sin n\theta_0|\ge c \gt 0$ where $\theta_0=(b-a)/2.$]

However, prove that the Fourier series converges at every point $x$.

I've computed the Fourier series and got $\frac{b-a}{2\pi}+\sum_{n\neq 0}\frac{e^{-ina}-e^{-inb}}{2\pi in}e^{inx}.$

Also, $|e^{-ina}-e^{-inb}|=2|\sin n\theta_0|$, and $\theta_0\in (0,\pi)$, so I can see that for infinitely many values of $n$, we have $|\sin n\theta_0|\ge c \gt 0$. But this does not guarantee $\sum_{n\neq 0}|\frac{e^{-ina}-e^{-inb}}{2\pi in}e^{inx}|\ge \sum \frac{c}{n}$, and in fact we might have this inequality only for the squares of integers, in which case the right hand side converges. So how does the hint solve the problem?

Moreover, for the second problem, to show that the Fourier series converges at every point, I think I need to use Dirichlet's test, using $1/n$ as the decreasing sequence to $0$, but how can I show that $\frac{e^{-ina}-e^{-inb}}{2\pi in}e^{inx}$ has bounded partial sums?

I would greatly appreciate any help.

Answer 1

Suppose $a \ne -\pi$ or $b \ne \pi$ and $a\ne b$. Then the function you are talking about must be discontinuous.

Suppose the series did converge absolutely for some $x$. That would mean $$ \sum_n \left|\frac{e^{-ina}-e^{-inb}}{2\pi in}\right| < \infty. $$ But that would force the uniform convergence of the Fourier series everywhere by the Weierstrass M-test. But uniform convergence would imply that the periodic extension of the limit function $\chi_{[a,b]}$ must be continuous everywhere, which only happens in the case that $a=-\pi$ and $b=\pi$, or $a=b$.

I'm not familiar with you text, but you should have some pointwise convergence theorem that shows the Fourier series converges to the mean of the left and right hand limits for your function.

Answer 2

Let us assume, by way of contradiction, that this series converges: $$\sum_{n\not=0} \frac{|e^{-na}+e^{-nb}|}{2\pi n}=\sum_{n\not=0}\frac{|\sin(n\theta_o)|}{\pi |n|}$$

We observe that due to our hypothesis that $-\pi < a < b \leq \pi$, the difference between $a$ and $b$ satisfies $0 < b - a < 2\pi$, which implies $0 < \theta_o < \pi$. When we state that the series converges, we are actually referring to taking the symmetric limit. This means that the partial sum denoted by $S_m$ converges:

$$S_m=\sum_{n=1}^m\frac{2|\sin(n\theta)|}{\pi n}$$

In particular the subsequence $S_{2m}$ but we may write this as:

$$S_{2m}=\sum_{n=1}^m \frac{2}{\pi }\left(\frac{|\sin(2n-1)\theta_o)|}{2n-1}+\frac{|\sin(2n\theta_o)|}{2n}\right)\geq \sum_{n=1}^m \frac{1}{\pi n}(|\sin(2n-1)\theta_o|+|\sin(2n\theta_o)|)$$

Let us investigate the conditions under which $g(x) = |\sin((2x-1)\theta_o)| + |\sin(2x\theta_o)| = 0$. This can only happen if both terms inside the sine functions are integer multiples of $\pi$, but this would imply that $\theta_o$ is a multiple of $\pi$, contradicting the specified range it belongs to. Consequently, $g(x)$ can never be zero. Since it has a periodicity of $\pi/\theta_o$, we can focus on the interval $[0, \pi/\theta_o]$, which is compact. We can then find the infimum of the function's range (which is achieved). Thus, we conclude that $g(x) > c$. Due to the periodic nature of $g(x)$, this lower bound must hold everywhere. As a result:

$$S_{2m}\geq \sum_{n=1}^m\frac{c}{\pi n}$$

This leads to an absurdity.

Answer 3

I see that both of the previous answers to part (b) of this exercise do not fully use the hint in the statement, so I'll attempt to do so. For part (c), the first few paragraphs of fonini's answer should suffice.

We need to show that the series $$ \sum_{n\ne0} \left| \frac{e^{-ina}-e^{-inb}}{2\pi in} e^{inx} \right| $$ converges. Rewriting the expression in terms of $\theta_0=(b-a)/2$ by factoring out $e^{-in\frac{1}{2}(a+b)}$ and recalling that $|e^{in\varphi}|=1$ for any $\varphi$ gives us $$ \sum_{n\ne0} \left| \frac{\sin n\theta_0}{\pi n} \right| = \frac{2}{\pi} \sum_{n=1}^\infty \frac{|\sin n\theta_0|}{n}, $$ since the value of the argument of the sum does not change if $n\to-n$.

Now comes the part of the proof where we use the 'for many values of $n$ one has …' part of the hint. To prove convergence, it suffices to show that in a countably infinite number of terms, $|\sin n\theta_0| \ge c > 0$, for if so, the sum of all these terms diverges by comparison with $\sum_{n=1}^\infty \frac{c}{n}$, which, in turn, implies divergence of our expression by comparison with this smaller series (insert zeroes in between the terms of the smaller series so that the identical terms in both series have the same index, i.e. match up).

To do this, we show that, for all integers $k>0$, we can choose another integer $n_k>0$ such that the desired property holds when $n=n_k$, dependent upon the value of $k$. First note that $0<\theta_0<\pi$, so there exists $0<c<1$ such that $\pi-\theta_0 < 2 \arcsin c$. (The choice of $2\arcsin c$ in favour of $\arcsin c$ is done to facilitate the key argument below.) We claim that, for each $k$, there exists a positive integer $n_k$ such that $$ n_k\theta_0 \in \big( (k-1)\pi + \arcsin c , k\pi - \arcsin c \big), $$ the interval on the right hand side we denote as $I_k$. For if so, we have $(k-1)\pi + \arcsin c < n_k\theta_0 < k\pi - \arcsin c$, which may be rearranged to give the desired condition. (Note that $|\sin(n_k\theta_0-(k-1)\pi)|=|\sin(n_k\theta_0-k\pi)|=|\sin(n_k\theta_0)|$.)

The claim can be proved as follows. (It helps to have a sketch of $y=|\sin x|$ on hand and draw on it as you read this argument.) Since $\theta_0<\pi$, it is obvious (by contradiction; simple exercise) that, for all $k$, each interval $((k-1)\pi,k\pi)$ includes at least one $n\theta_0$. Now there are two cases: either $n\theta_0 \in I_k$ or $n\theta_0 \notin I_k$. In the first case, we are done, and can set $n_k=n$. In the second case, either $n\theta_0 \in ((k-1)\pi, (k-1)\pi+\arcsin c)$ or $n\theta_0 \in (k\pi-\arcsin c, k\pi)$. But since $|I_k|>n_k\theta_0$, in the first subcase $(n+1)\theta_0 \in I_k$ (so $n_k=n+1)$, whereas in the second subcase $(n-1)\theta_0 \in I_k$ (so $n_k=n-1$), and we are also done. (Now you should be able to see why we chose $2\arcsin c$.) Therefore, each interval $((k-1)\pi,k\pi)$ (the number of which is countably infinite) can be assigned to an integer $n_k$, so we conclude that there are infinitely many $n_k$, such that if $n$ does not satisfy our desired condition, then either $n-1$ or $n+1$ will, i.e. there cannot be consecutive values of $n$ which do not satisfy the condition. We conclude that, in the worst case scenario where the $n_k$ are most sparsely distributed, either the $n_k$ are all the odd positive integers or $n_k$ are all the even positive integers. If we take the reciprocal of each and sum over each collection separately, both diverge (since they are related to the harmonic series).

Thus there exists an infinite number of $n_k$ such that $|\sin n_k\theta_0| \ge c$, enumerated by the positive integers $k$. This implies that $$ \sum_{n=1}^\infty \frac{|\sin n\theta_0|}{n} \ge \sum_{k=1}^\infty \frac{|\sin n_k\theta_0|}{n_k} \ge \sum_{n\,\text{odd}} \frac{c}{n} \to +\infty, $$ by the arguments presented in paragraph 3. We are done.

[Note: The book claims to assume very few prerequisites, so in all candour, I don't know why the authors assigned this as Exercise 9(b) in a collection of 20 exercises without more hints, since this argument is somewhat intricate.]